jackiem1075 said:
Thanks we have put something together. We did the velocity of the ping pong when it lands on earth
We found out the time it took to fall and then did time x acceleration due to gravity = Velocity
for time we did height(3m) x 2 / 9.81 = 0.6116207951 then found the square root of that answer = 0.782 seconds
We then did 0.782 x 9.81 = 7.672 m/s
Is this right for a basic level?
It gives the right answer and is a correct approach, so that is good.
I have two suggestions to make.
1. It is good to work with symbols and get the result with algebra rather than numbers.
In the case at hand, you start by trying to determine time. You know distance and acceleration. The standard symbols for those quantities are "t" for time, "s" for distance or displacement and "a" for acceleration. The equation you would have started with would be:$$s=\frac{1}{2}at^2$$You solved that correctly (multiply both sides by 2, divide both by a, take the square root of both and then flip left with right) to arrive at$$t=\sqrt{2\frac{s}{a}}$$One assumes that your child did all of this correctly. The next step would to multiply acceleration (a) by time (t) to arrive at the change in velocity (##\Delta v##):$$\Delta v=at$$But since we already have the formula for t above, we can substitute that in, yielding:$$\Delta v = a \sqrt{2\frac{s}{a}}$$
This is where doing things with symbols starts to shine compared to doing things with numbers. You can cancel that ##\sqrt{a}## in the denominator with the ##a## in the numerator yielding a single ##\sqrt{a}## in the numerator:$$\Delta v = \sqrt{2as}$$Now with your desired quantity alone on the left hand side, you can plug numbers into the right hand side, evaluate and done:$$\Delta v =\sqrt{2as} = \sqrt{2 \times 9.81 \times 3} = 7.672 \frac{m}{s}$$That is the first point I wanted to make: Try to use symbols instead of numbers.2. There is another approach that could have been taken. An "energy" approach was suggested. This time the starting point is:$$PE_{\text{initial}}=KE_{\text{final}}$$You plug in your formulas for potential energy (##PE=mgh##) and kinetic energy (##KE=\frac{1}{2}mv^2##) and get:$$mgh=\frac{1}{2}mv^2$$Now you can divide both sides by m (eliminating m entirely), multiply by 2, take the square root and flip left and right yielding:$$v=\sqrt{2gh}$$By no coincidence, this is the exact same result that we had obtained above. It is just that we are using "##g##" for gravity instead of "##a##" for acceleration and "##h##" for height instead of "##s##" for displacement.
There is actually a third approach that could have been taken. It is closely related to the "energy" approach. This time the starting point is the "SUVAT" equations. This is a set of equations that relate displacement (##s##), starting velocity (##u##), final velocity (##v##), acceleration (##a##) and time (##t##) in various arrangements. You can Google for "SUVAT" and find a bazillion hits. If one does so and searches for an equation with ##v## on the left hand side and variables you have as givens on the right, one quickly arrives at:$$v^2=u^2+2as$$ Substitute in starting velocity (##u##) = 0, acceleration (##a##)=9.81, displacement (##s##) = 3. Take the square root and you are done.
